Strain energy due to bending, rectangular beam

Strain energy due to bending is the energy stored in a beam because of bending (flexural) stresses.

Derivation

Consider an elemental length $dx$ of a beam under a bending moment $M$. The ends of the element rotate relative to each other by

$$d\theta = \frac{M\,dx}{EI}$$

The moment rises gradually from 0 to $M$ while the rotation rises from 0 to $d\theta$, so the work done (strain energy stored) in the element is

$$dU = \tfrac12 M\,d\theta = \frac{M^2\,dx}{2EI}$$

Alternative (stress method): the bending stress at a distance $y$ from the neutral axis is $\sigma = \dfrac{My}{I}$, and the energy per unit volume is $\dfrac{\sigma^2}{2E}$:

$$dU = \int \frac{M^2y^2}{2EI^2}\,dA\,dx = \frac{M^2}{2EI^2}\left(\int y^2\,dA\right)dx = \frac{M^2\,dx}{2EI}$$

For the whole beam of length $L$:

$$U_b = \int_0^L \frac{M^2}{2EI}\,dx$$

For a rectangular beam

With $I = \dfrac{bd^3}{12}$, a simply supported beam of span $L$ carrying a central load $P$ has $M = \dfrac{P x}{2}$ for $0 \le x \le L/2$:

$$U_b = 2\int_0^{L/2}\frac{P^2x^2}{8EI}\,dx = \frac{P^2L^3}{96EI} = \frac{P^2L^3}{8Ebd^3}$$

For a cantilever with an end load $P$: $M = Px$, so $U_b = \dfrac{P^2L^3}{6EI} = \dfrac{2P^2L^3}{Ebd^3}$.

Result: $U_b = \displaystyle\int_0^L \frac{M^2}{2EI}\,dx$